In vinyl acetylene $CH \equiv C - CH = C{H_2}$ type of overlapping in $\left( {{C_2}\sigma {C_3}} \right)$ bond is
A.$s{p^2} - sp$
B.$sp - s{p^2}$
C.$s{p^3} - s{p^3}$
D.$s{p^3} - s{p^2}$
Answer
Verified
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Hint:We have to take into consideration the number of bonds formed by each carbon and determine whether the bonds are $\sigma - $bonds (or) $\pi - $bonds. We can determine the hybridization of each atom of carbon using the $\sigma - $bonds (or) $\pi - $bonds.
Complete step by step answer:
We have to first identify the name of atoms of carbon. We can give the structure as,
\[\mathop C\limits_1 H \equiv \mathop C\limits_2 - \mathop C\limits_3 H = \mathop C\limits_4 {H_2}\]
We have to find the hybridization of two carbons present in the middle. Let us carbon-2.
From the written structure, we could see that it is forming one $\sigma - $bond and two $\pi - $bonds with carbon-3. We can also see it does not form any bonds with hydrogen atoms.
Therefore, it contains two $\sigma - $bonds. We have to consider the bond pairs while determining the hybridization of the atoms. We should not consider the $\pi - $bonds while determining the hybridization of the atom.
We can say that there are two bond pairs involved, and therefore, 2 hybridized orbitals are needed to accommodate them. Hence, the hybridization of this carbon atom will be $sp$.
Let us now consider carbon-3.
From the structure, we could see that one $\sigma - $bond and two $\pi - $bonds with carbon-2. We can also see it forms one $\sigma - $bond with a hydrogen atom.
Let us consider the number of $\sigma - $bond to calculate the hybridization of the carbon. There are 3 $\sigma - $bonds present and 3 hybridized orbitals are needed to accommodate 3 bond pairs. Therefore, the hybridization of the carbon atom will be $s{p^2}$.
Therefore, the option (B) is correct.
Note: We have to be careful while calculating the number of $\sigma - $bond which is formed by each of the carbon and the hydrogen atom since they are not explicitly shown and there is a chance one might miss them. We can prepare vinyl acetylene by Hofmann elimination of the quaternary ammonium salt. We can also prepare using dehydrohalogenation of 1,3-dichloro-2-butene.
Complete step by step answer:
We have to first identify the name of atoms of carbon. We can give the structure as,
\[\mathop C\limits_1 H \equiv \mathop C\limits_2 - \mathop C\limits_3 H = \mathop C\limits_4 {H_2}\]
We have to find the hybridization of two carbons present in the middle. Let us carbon-2.
From the written structure, we could see that it is forming one $\sigma - $bond and two $\pi - $bonds with carbon-3. We can also see it does not form any bonds with hydrogen atoms.
Therefore, it contains two $\sigma - $bonds. We have to consider the bond pairs while determining the hybridization of the atoms. We should not consider the $\pi - $bonds while determining the hybridization of the atom.
We can say that there are two bond pairs involved, and therefore, 2 hybridized orbitals are needed to accommodate them. Hence, the hybridization of this carbon atom will be $sp$.
Let us now consider carbon-3.
From the structure, we could see that one $\sigma - $bond and two $\pi - $bonds with carbon-2. We can also see it forms one $\sigma - $bond with a hydrogen atom.
Let us consider the number of $\sigma - $bond to calculate the hybridization of the carbon. There are 3 $\sigma - $bonds present and 3 hybridized orbitals are needed to accommodate 3 bond pairs. Therefore, the hybridization of the carbon atom will be $s{p^2}$.
Therefore, the option (B) is correct.
Note: We have to be careful while calculating the number of $\sigma - $bond which is formed by each of the carbon and the hydrogen atom since they are not explicitly shown and there is a chance one might miss them. We can prepare vinyl acetylene by Hofmann elimination of the quaternary ammonium salt. We can also prepare using dehydrohalogenation of 1,3-dichloro-2-butene.
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